Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in PYTHON):

Solution:
###modular exponentiation: a^n % M
#pow(a, n, M) 
### gcd: gcd(a, b)
#import fractions
#fractions.gcd(a, b)
### LCM (M, N)
#lcm = M*N/fractions.gcd(M, N)

def get_res_row(L, n, m):
    result = 0
    res_txt = ''
    Lrt = []
    rcnt = 0
    for row in L:
        rcnt += 1
        mr = min(row)
        x = []
        for elem in row:
            x.append(elem-mr)
        result += mr
        if mr > 0: 
            res_txt += ('row ' + str(rcnt) + '\n')*mr
        Lrt.append(x)
    for c in range(m):
        col = [Lrt[y][c] for y in range(n)]
        mc = min(col)
        for r in range(n):
            Lrt[r][c] -= mc
        result += mc
        if mc > 0:
            res_txt += ('col ' + str(c+1) + '\n')*mc
    for row in Lrt:
        if max(row) != 0:
            result = -1
            break

    return (result, res_txt[:-1])

def get_res_col(L, n, m):
    result = 0
    res_txt = ''
    Lrt = []
    rcnt = 0
    for c in range(m):
        col = [L[y][c] for y in range(n)]
        mc = min(col)
        for r in range(n):
            L[r][c] -= mc
        result += mc
        if mc > 0:
            res_txt += ('col ' + str(c+1) + '\n')*mc

    for row in L:
        rcnt += 1
        mr = min(row)
        x = []
        for elem in row:
            x.append(elem-mr)
        result += mr
        if mr > 0: 
            res_txt += ('row ' + str(rcnt) + '\n')*mr
        Lrt.append(x)

    for row in Lrt:
        if max(row) != 0:
            result = -1
            break

    return (result, res_txt[:-1])

#N = int(raw_input())  
n, m = [int(s) for s in raw_input().split(" ")]
L = []
for i in range(n):
    x = [int(s) for s in raw_input().split(" ")]
    L.append(x)
if n <= m:
    result, res_txt = get_res_row(L, n, m)
else:
    result, res_txt = get_res_col(L, n, m)
if result <= 0:
    print "{}".format(result)
else:
    print "{}".format(result)
    print "{}".format(res_txt)